From: baez@galaxy.ucr.edu (John Baez) Subject: Re: Geometric quantization Date: 12 Jul 2000 03:05:04 GMT Newsgroups: sci.physics.research In article <8katdc$qcd$1@bob.news.rcn.net>, Michael Weiss wrote: >There are a bunch of things I don't understand about geometric >quantization, at least not in the way I'd *like* to understand them. Yes, let's talk about this stuff more! Toby and I have been busy wrestling the first Chern class to the ground, but we're almost done with that - and then we can get back to geometric quantization per se. Pretty soon I'll post an outline sketching what geometric quantization is supposed to do. That may help you see the "big picture". But for now, a few questions: >a) Why does the Kostant-Kirillov symplectic form scale as 1/R instead of >1/R^2? I explained this in a reply to Alex Katz (see the end of this post). Let me know if you don't like this explanation. >a+1/2) How do you spell Kirillov? The way you just wrote it - after I corrected what you wrote. :-) >b) What does co-multiplication look like on so(3)*? ("Look" as in mental >video.) Mental videos, he wants! What do you think this is, a video store? I know what the multiplication on so(3) looks like: [-,-] : so(3) x so(3) -> so(3) It's just the cross product we all learned about in school! We can think of it as a linear map if we like m: so(3) tensor so(3) -> so(3) and then we can dualize it to get the comultiplication on so(3)*: m*: so(3)* -> so(3)* tensor so(3)* It's still just the cross product in disguise, so if I were forced to visualize it, I would just visualize the cross product while standing on my head and chanting "dualization, dualization, dualization...." But you wanna know what the comultiplication is LIKE? Well, the map m looks like this: i tensor i -> 0 j tensor j -> 0 k tensor k -> 0 i tensor j -> k j tensor k -> i k tensor i -> j j tensor i -> -k k tensor j -> -i i tensor k -> -j (I just wrote down a bunch of cross products using some fancy lingo.) So we get the map m* by taking the adjoint of m: i -> j tensor k - k tensor j j -> k tensor i - i tensor k k -> i tensor j - j tensor i Okay, now I see what this "looks like"! It looks like the map sending each vector to the infinitesimal rotation about that vector. Or to the 2-form corresponding to the oriented plane orthogonal to that vector. (In each description here I am committing the sin of identifying a vector space with its dual - but this is no sin, since we needed an inner product to define so(3) in the first place.) But beware: this mental video only applies to so(3)! If we were doing so(42) we wouldn't be able to pull these tricks - only in 3 dimensions can we flip-flop back and forth between vectors and infinitesimal rotations as I have been doing here. >c) Why does the topological definition of the Chern number agree with the >connection definition? Well, Toby worked this out in detail, but I guess that was *too much* detail for you, so I'll give you something vague: The topological definition of the Chern number tells us how to build a line bundle over the surface X with Chern number n. We start with the trivial line bundle over X. Then we cut out a little disk from X. Then we glue the disk back in, and glue the line bundle back together - but with a twist! How much of a twist? Simple: to glue the line bundle back together, we need a map from the boundary of the disk to U(1). That's a map from the circle to the circle. So, just make sure this map has winding number n. Then we'll get a bundle with Chern number n. Okay, now for the connection definition! Here we pick any connection on our line bundle, compute its curvature, integrate it over the surface X, and divide by 2 pi i. Why do these definitions agree? Well, let's take the simplest possible connection on the line bundle we just built. We can take this connection to be the obvious trivial flat connection everywhere *except* on that little disk. But it can't be flat on that disk - unless n = 0. Why? Well, if you think about it, when we parallel transport a vector around the edge of the disk, it has to turn around n times. In other words, if we describe our U(1) connection by u(1)-valued 1-form A, we must have integral_{boundary of disk} A = 2 pi i n But by Stokes' theorem we have integral_{boundary of disk} A = integral_{disk} F where F = dA is the curvature of A. So we must have integral_{disk} F = 2 pi i n But the connection is flat except on the disk, so we have integral_X F = 2 pi i n Voila! Now, I used various clever little observations in the process of carrying out this argument, and I have done my darndest to sweep most of them under the rug where you won't see them - so if you see any funny lumps in the carpet, don't worry. >d) What does the covariant derivative look like on the vector bundles we >cooked up, and why isn't it holomorphic? "Look like"? Mental videos he wants! Let me explain why U(1) connections are rarely holomorphic. Let's work locally, so a U(1) connection is just a 1-form taking values in the imaginary numbers. (Or real numbers, if you're a physicist - just divide by i if you prefer this.) Now, for A to be holomorphic means that A = A_z dz = A_z (dx + idy) where A_z is a holomorphic function. But wait a minute! A minute ago I said A had to be a 1-form taking values in the imaginary numbers: that means A = A_x dx + A_y dy where A_x and A_y are imaginary-valued functions. If you compare the 2 equations I just wrote, you'll see the functions A_z and A_x are the same, and that this function must be HOLOMORPHIC yet PURELY IMAGINARY. But there *aren't any* holomorphic purely imaginary functions except constant functions! So U(1) connections are rarely holomorphic - just like real-valued (or imaginary-valued) functions are rarely holomorphic. If you want a holomorphic connection you gotta use the gauge group C* (the nonzero complex numbers) instead of U(1). But for physics purposes there are lots of reasons why we really want the gauge group to be U(1) when we're doing geometric quantization. We want a well-defined HILBERT SPACE of sections at the end of the day, after all! That means gauge transformations can only multiply sections by a phase - not by the number 42, say. So there is a kind of tension between our desire to work with U(1) and our desire to cook up a Hilbert space of holomorphic sections. But tension is good: it keeps the soap opera going.... stay tuned! .............................................................................. From: baez@galaxy.ucr.edu (John Baez) Subject: Re: Geometric quantization, Question for John Date: 02 Jun 2000 00:00:00 GMT In article <200005250055.RAA25412@king.halcyon.com>, wrote: >(When you have time from moving; though I don't believe that anyone who >just moved into a new house and must have even more books than me has >time for all these newsgroup postings...) I don't know how many you have, but I don't actually think I have very many books - at least compared to my friends. So by now I've moved in, and when I'm not sitting in the hot tub watching the sun set, I spend my evenings relaxing by posting lots of articles to s.p.r. - as you can probably see. >I read the whole paper and really liked it. Great, thanks! >However, there's a key (probably simple) question I have: > >On the equation at the end of page 8 (how come no equation numbers??) >[concerning SO(3) and the identification of the dual space with R^3 > and the dual paring <.,.> becomes the Euclidean inner product...] : > > {l,m}(x) = (I understand this) > >and so the two form is: > > omega(a,b) = 1/|x|^2 (I DON'T understand this) > >Where does the 1/|x|^2 come from? and why is it different (by this >factor) from the previous line?? [notation slightly modified by jb] Well, it actually took me a while to remember where this factor comes from, but it goes sort of like this. In the first formula we are working with elements of the Lie algebra so(3) called l and m, while in the second formula we are working with tangent vectors to the sphere of radius r = |x| called a and b. So the issue is how we translate from one to the other. Now I assume you know that an element of so(3) is just a 3x3 antisymmetric matrix, but that can we think of it an "infinitesimal rotation" on R^3 - which is really a special sort of vector field on R^3. For example, the 3x3 antisymmetric matrix 0 -1 0 1 0 0 0 0 0 corresponds to an infinitesimal rotation about the z axis, which is the vector field (-y, x, 0) on R^3. Such vector fields are tangent to the sphere of radius r centered at the origin - no matter what r is. That's how we translate from elements of so(3) to vector fields tangent to the sphere of radius r. And notice that these vector fields get *bigger* as r gets bigger - their magnitude is proportional to r. For example, on the sphere of radius 1 the magnitude of (-y, x, 0) is at most 1, while on the sphere of radius 2 its magnitude is at most 2, and so on. That's where the factor of 1/|x|^2 comes from in the formula you don't understand: when we take the tangent vectors a and b and find Lie algebra elements l and m corresponding to them, we pick up two factors of 1/r, where r = |x| is the radius of the sphere. What I just gave you here is more of an "intuitive explanation" than a "proof", so if you want me to do a calculation proving that the 1/|x|^2 must appear in this formula, just ask. The paper has a couple of paragraphs before this formula that provide the machinery to do this calculation. >I (think) I understand the rest; if the above is true, the integral >over a symplectic leaf is +- 4pi R, and in (pre)quantizing must >be a multiple of 2pi, so j is half integer, etc. Right.